How do you find the domain and range of #f(x) = sqrt(4 - x²)#?

1 Answer
Jul 2, 2015

Answer:

Domain: #-2<=x<=2#;
Range: #0<=y<=2#

Explanation:

You want to avoid to have a negative number as argument of your square root, so you must set:
#4-x^2>=0# that can be rearranged to get (changing signs):
#x^2<=4#
#x<=+-sqrt(4)#
#x<=+-2#
So, basically, your function can accept only #x# values inside the interval from #-2# to #+2# giving a domain: #-2<=x<=2#.
The maximum #y# value of your function is reached when #x=0# and corresponds to #y=2#. So the range will be from #0# to #+2# or: #0<=y<=2#

Graphically:
graph{sqrt(4-x^2) [-10, 10, -5, 5]}